Mathematics, 30.11.2019 03:31 aubreykenzie686

Minor surgery on horses under field conditions requires a reliable short-term anesthetic producing good muscle relaxation, minimal cardiovascular and respiratory changes, and a quick, smooth recovery with minimal aftereffects so that horses can be left unattended. an article reports that for a sample of n = 69 horses to which ketamine was administered under certain conditions, the sample average lateral recumbency (lying-down) time was 18.94 min and the standard deviation was 8.3 min. does this data suggest that true average lateral recumbency time under these conditions is less than 20 min? test the appropriate hypotheses at level of significance 0.10.state the appropriate null and alternative hypotheses.

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Answer from: kat9940

Null hypothesis: $\mu \geq 20$

Alternative hypothesis: $\mu < 20$

$P(t_{68}$

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v\alpha$ so we can conclude that we FAIL reject the null hypothesis, and the the actual sample average lateral recumbency is not significantly lower than 20.

Step-by-step explanation:

1) Data given and notation

$\bar X=18.94$ represent the mean for the sample

s=8.3 represent the standard deviation for the sample

n=69 sample size

$\mu_o =20$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

p_v represent the p value for the test (variable of interest)

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the true average lateral recumbency time under these conditions is less than 20 min:

Null hypothesis: $\mu \geq 20$

Alternative hypothesis: $\mu < 20$

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

3) Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{18.94-20}{\frac{8.3}{\sqrt{69}}}=-1.06$